5 In this question you must show detailed reasoning. Fig. 12 shows part of the graph of \( y=x^{2}+\frac{1}{x^{2}} \). The tangent to the curve \( y=x^{2}+\frac{1}{x^{2}} \) at the point \( \left(2, \frac{17}{4}\right) \) meets the \( x \)-axis at A and meets the \( y \)-axis at B . Ois the origin. Find the exact area of the triangle OAB .

To find the area of triangle OAB formed by the origin O, point A where the tangent meets the x-axis, and point B where the tangent meets the y-axis, we will follow these steps: 1. **Find the derivative of the function** to determine the slope of the tangent line at the point \( \left(2, \frac{17}{4}\right) \). 2. **Use the point-slope form of the equation of the tangent line** to find the equations of the lines that meet the axes. 3. **Determine the coordinates of points A and B**. 4. **Calculate the area of triangle OAB** using the coordinates of points O, A, and B. ### Step 1: Find the derivative The function is given by: \[ y = x^2 + \frac{1}{x^2} \] To find the derivative \( y' \): \[ y' = \frac{d}{dx}(x^2) + \frac{d}{dx}\left(\frac{1}{x^2}\right) = 2x - \frac{2}{x^3} \] Now, we evaluate the derivative at \( x = 2 \): \[ y'(2) = 2(2) - \frac{2}{(2)^3} = 4 - \frac{2}{8} = 4 - \frac{1}{4} = \frac{16}{4} - \frac{1}{4} = \frac{15}{4} \] So, the slope of the tangent line at the point \( \left(2, \frac{17}{4}\right) \) is \( \frac{15}{4} \). ### Step 2: Equation of the tangent line Using the point-slope form of the equation of a line: \[ y - y_1 = m(x - x_1) \] where \( (x_1, y_1) = \left(2, \frac{17}{4}\right) \) and \( m = \frac{15}{4} \): \[ y - \frac{17}{4} = \frac{15}{4}(x - 2) \] Multiplying through by 4 to eliminate the fraction: \[ 4y - 17 = 15(x - 2) \] Expanding this: \[ 4y - 17 = 15x - 30 \] Rearranging gives: \[ 4y = 15x - 13 \quad \Rightarrow \quad y = \frac{15}{4}x - \frac{13}{4} \] ### Step 3: Find points A and B **Point A (x-axis intersection)**: Set \( y = 0 \): \[ 0 = \frac{15}{4}x - \frac{13}{4} \] Solving for \( x \): \[ \frac{15}{4}x = \frac{13}{4} \quad \Rightarrow \quad 15x = 13 \quad \Rightarrow \quad x = \frac{13}{15} \] Thus, point A is \( \left(\frac{13}{15}, 0\right) \). **Point B (y-axis intersection)**: Set \( x = 0 \): \[ y = \frac{15}{4}(0) - \frac{13}{4} = -\frac{13}{4} \] Thus, point B is \( \left(0, -\frac{13}{4}\right) \). ### Step 4: Calculate the area of triangle OAB The vertices of triangle OAB are: - O: \( (0, 0) \) - A: \( \left(\frac{13}{15}, 0\right) \) - B: \( \left(0, -\frac{13}{4}\right) \) The area \( A \) of triangle OAB can be calculated using the formula: \[ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \] Substituting the coordinates: \[ \text{Area} = \frac{1}{2} \left| 0\left(0 + \frac{13}{4}\right) + \frac{13}{15}\left(-\frac{13}{4} - 0\right) + 0\left(0 - 0\right) \right| \] This simplifies to: \[ = \frac{1}{2} \left| \frac{13}{15} \left(-\frac{13}{4}\right) \right| = \frac{1}{2} \cdot \frac{13 \cdot 13}{15 \cdot 4} = \frac{169}{120} \] Thus, the exact area of triangle OAB is: \[ \boxed{\frac{169}{120}} \]